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this question must be pretty easy but i´m just taking my firs course in complex variables, i need to find the limit of Z over it conjugate as z reach the infinity. I need to know how to do it in a rigorous way and informal way, so i need your help. By formal, i mean with neighborhood and epsilon definition of limit.

I have tried with trig form of Z, but i don´t reach any answer. Thanks

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Please be careful with your variables. It appears Z and z are the same thing, so they should get the same name. – Ross Millikan Aug 14 '12 at 4:21
If you want a really simple example, (i) Let $z=r$ where $r$ is huge; (ii) let $z=ri$ where $r$ is huge. In one case, ratio is $1$, in the other $-1$. Now for formal proof. Let $a$ be the supposed limit, and let $\epsilon=\frac{3}{4}$. Then however large $r$ is, there are $z$ with $|z|\gt r$ such that $|z/\bar{z}-a|\gt \epsilon$, since no complex number can be simultaneously within $3/4$ of $1$ and within $3/4$ of $-1$ (Triangle Inequality). – André Nicolas Aug 14 '12 at 4:48
up vote 4 down vote accepted

Hint: The trigonometric form is a good idea. What is $\dfrac{r(\cos\theta+i\sin\theta)}{r(\cos\theta-i\sin\theta)}$ when $r$ is large? Does it depend on $\theta$?

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You're missing a parenthesis. The \dfrac is to improve readability. For me, and I guess for many ohters, the letters are kind of tiny. – Pedro Tamaroff Aug 14 '12 at 4:15
@PeterTamaroff: My mild edit overwrote yours. Can you edit again? – André Nicolas Aug 14 '12 at 4:16

We have $z\bar{z}=|z|^2$ so that $z/\bar{z}=z^2/|z|^2=\big(z/|z|\,\big)^2$, which very much depends on $\arg z$.

Indeed, picking any $M>0$ and $s$ with $|s|=1$ we can select a $z$ with $|z|=M$ and $(z/|z|)^2=s$; can you see why this should be true algebraically and geometrically? This means every value on the unit circle is obtained somewhere in every neighborhood of $\infty$; can all of the distances between points on the unit circle be made as small as we wish?

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Perhaps even easier than the trig form is the exponential form. If $z=re^{i\theta}$, then $\bar z=re^{-i\theta}$, so you’re looking at the limit of $$\frac{re^{i\theta}}{re^{-i\theta}}=e^{2i\theta}$$ as $r$ gets large.

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If $z=r e^{i \theta}$, then $\frac{z}{\overline{z}} = e^{i 2 \theta}$.

If I choose $r_n = n$, and $\theta_n = n \frac{\pi}{2}$, then with $z_n=r_n e^{i \theta_n}$, we have $z_n \to \infty$ and $\frac{z_n}{\overline{z_n}} = (-1)^n$, consequently the ratio has no limit.

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